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Chapter 17: Problem 103

A police car is chasing a speeding Porsche 911 . Assume that the Porsche's maximum speed is \(80.0 \mathrm{~m} / \mathrm{s}\) and the police car's is \(54.0\) \(\mathrm{m} / \mathrm{s}\). At the moment both cars reach their maximum speed, what frequency will the Porsche driver hear if the frequency of the police car's siren is \(440 \mathrm{~Hz}\) ? Take the speed of sound in air to be \(340 \mathrm{~m} / \mathrm{s}\).

Short Answer

Expert verified
The Porsche driver hears a frequency of approximately 469 Hz.

Step by step solution

01

Identify Known Values

We have the following values: \( v_{s} = 54 \text{ m/s} \) (speed of the source, the police car), \( v_{o} = 80 \text{ m/s} \) (speed of the observer, the Porsche), \( f = 440 \text{ Hz} \) (frequency of the siren), and \( v = 340 \text{ m/s} \) (speed of sound in air).
02

Choose Doppler Effect Formula

As the observer (Porsche) is moving away from the source (police car), we'll use the Doppler effect formula:\[ f' = f \left( \frac{v + v_{o}}{v + v_{s}} \right) \] where \( f' \) is the observed frequency.
03

Substitute the Known Values into the Formula

Substitute the given values into the Doppler effect formula: \[ f' = 440 \left( \frac{340 + 80}{340 + 54} \right) \] Calculate the expression inside the parentheses.
04

Calculate the Observed Frequency

Calculate the terms inside the fraction:\(\frac{340 + 80}{340 + 54} = \frac{420}{394} \approx 1.06599 \).Now compute \( f' = 440 \times 1.06599 \approx 469 \text{ Hz} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Frequency
Frequency is a fundamental concept in wave mechanics and acoustics. It refers to the number of times a wave repeats itself in a second. Frequency is measured in hertz (Hz), where 1 Hz equals one cycle per second. It's a crucial component when discussing sounds, such as sirens or musical notes.

In the context of the Doppler Effect, frequency is what changes when either the source of the sound or the listener is moving. For example, if a police car with a siren is approaching you, the sound of the siren appears to have a higher frequency than when it is moving away. This change in frequency due to relative motion is what defines the Doppler Effect.

Understanding frequency not only helps in sound analysis but also in analyzing motion and speed in physics. It connects deeply with how we perceive different pitches and tones in the environment.
Wave Motion
Wave motion refers to the transfer of energy and momentum between points through oscillations, without the transport of matter. Waves can be seen in water, sound, and even light. Sound waves, like those from a siren, are mechanical waves that spread through mediums like air.

Sound waves are longitudinal waves where particles of the medium vibrate parallel to the direction of the wave. When a police car siren blares, sound waves travel through the air to reach our ears or the ears of someone in a vehicle like the Porsche. Understanding wave motion helps us comprehend how sound travels and why it might change as we or the source move.

Wave motion allows sound to be heard over distances and explains phenomena like echoes, resonance, and the Doppler Effect, which all depend on wave interactions and the medium they travel through.
Speed of Sound
The speed of sound is the rate at which sound waves travel through a medium. In air, this speed is approximately 340 m/s at room temperature. However, it can vary based on medium, temperature, and atmospheric pressure.

In our exercise, this speed plays a critical role. It is used in the Doppler effect formula to calculate how frequencies change with movement. The speed of sound is a constant that helps in converting changes in relative velocity between a sound source and observer into perceived changes in frequency.

Knowing how fast sound travels is essential for various practical applications. It is used in meteorology, sonar, and wireless communications. For example, sound travels faster in water than through air, which is crucial for submarine navigation and detecting underwater objects.
Relative Motion
Relative motion is the motion of one object with respect to another. In the Doppler Effect, it particularly concerns the movement of the sound source and the observer.

In our given scenario, the relative motion is between the moving police car and the speeding Porsche. The observed frequency of the siren changes because the Porsche is moving away from the police car, increasing the distance between them. This increasing distance results in a lower apparent frequency.

Understanding relative motion is crucial because it helps explain many everyday occurrences, such as why a car's engine sounds different when it drives past us, and helps us calculate the effects on perceived sound accurately, which is quite useful in various technological and scientific fields.

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Most popular questions from this chapter

A source \(S\) and a detector \(D\) of radio waves are a distance \(d\) apart on level ground (Fig. \(17-52\) ). Radio waves of wavelength \(\lambda\) reach D either along a straight path or by reflecting (bouncing) from a certain layer in the atmosphere. When the layer is at height \(H\), the two waves reaching \(\mathrm{D}\) are exactly in phase. If the layer gradually rises, the phase difference between the two waves gradually shifts, until they are exactly out of phase when the layer is at height \(H+h\). Express \(\lambda\) in terms of \(d, h\), and \(H\).

Ultrasound, which consists ultrasour of sound waves with frequencies above the human audible range, can Artery be used to produce an image of the interior of a human body. Moreover, ultrasound can be used to measure Figure 17-47 Problem \(83 .\) the speed of the blood in the body; it does so by comparing the frequency of the ultrasound sent into the body with the frequency of the ultrasound reflected back to the body's surface by the blood. As the blood pulses, this detected frequency varies. Suppose that an ultrasound image of the arm of a patient shows an artery that is angled at \(\theta=20^{\circ}\) to the ultrasound's line of travel (Fig. 17-47). Suppose also that the frequency of the ultrasound reflected by the blood in the artery is increased by a maximum of \(5495 \mathrm{~Hz}\) from the original ultrasound frequency of \(5.000000 \mathrm{MHz} .\) (a) In Fig. \(17-47\), is the direction of the blood flow rightward of leftward? (b) The speed of sound in the human arm is \(1540 \mathrm{~m} / \mathrm{s}\). What is the maximum speed of the blood? (Hint: The Doppler effect is caused by the component of the blood's velocity along the ultrasound's direction of travel.) (c) If angle \(\theta\) were greater, would the reflected frequency be greater or less?

You are standing at a distance \(D\) from an isotropic point source of sound. You walk \(50.0 \mathrm{~m}\) toward the source and observe that the intensity of the sound has doubled. Calculate the distance \(D\).

An avalanche of sand along some rare desert sand dunes can produce a booming that is loud enough to be heard \(10 \mathrm{~km}\) away. The booming apparently results from a periodic oscillation of the sliding layer of sand-the layer's thickness expands and contracts. If the emitted frequency is \(90 \mathrm{~Hz}\), what are (a) the period of the thickness oscillation and (b) the wavelength of the sound?

Two trains are traveling toward each other at \(30.5 \mathrm{~m} / \mathrm{s}\) relative to the ground. One train is blowing a whistle at \(500 \mathrm{~Hz}\). (a) What frequency is heard on the other train in still air? (b) What frequency is heard on the other train if the wind is blowing at \(30.5 \mathrm{~m} / \mathrm{s}\) toward the whistle and away from the listener? (c) What frequency is heard if the wind direction is reversed?
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